Parabolas ( making equation )

Question

A parabola with vertex (1,5) has x-intercepts (a, 0) and (b, 0) where a<0, b>0, and 3 |a|=b . Find a·b.

I tried making intercept form but it I can't do it. Please help.

Answer 1

You can start with a formula that involves the vertex $(h,k)$:

$$(x-h)^2 = -K(y-k).$$

(I'm assuming vertical axis of symmetry.)

In this case you know the vertex:

$$(x-1)^2 = -K(y-5) \\ x^2 - 2x + (1 - 5K) = -Ky.$$

The solutions for $y=0$ are

$$x = \frac{2 \pm \sqrt{4 - 4(1 - 5K)}}{2} = 1 \pm \sqrt{5K} = 1 \pm C,$$

where $C$ is a constant.

If the two solutions are $(a,0), (b,0)$ with $a$ negative and $b$ positive, and $3|a| = b$, then we see that $C=2$, and $a \cdot b = -1 \cdot 3 = -3$.